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Lkl <br />CURVE TABLES. <br />Published by KEUFFEL 8v ESSER CO. <br />`H®W TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext, to any other radius may be f bund nearly enough, by dividing the Tan. <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />' To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />To find `Deg. of Curve, having the Central Angle and External: <br />Divide Ext.. opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Est. of twice the given angle divided by the radius of a 1° -curve will <br />be. the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE. <br />Wanted a. Curve with an Ext. of about 12 ft. Angle <br />of Intersectiori or I. P.=231 20' to the R. at Station <br />542-x-72. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87--.12=10.07. Say a lOP Curve. <br />Tan. in Tab. I opp. 23° 20'= 1183.1 <br />1183.1=10 —118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31 x-0.16,=118.47 =corrected Tangent. <br />(If corrected Ext, is required find in same way) <br />Ang.23°20'=23.33° : 10=2.3333=L. C. <br />2°19=def. for star 542 I.P.=sta. 542-x-72, <br />4049 It _ " " " +50. Tan. = 1 .18.47 <br />70 1911= It It " 543 <br />° B. C. = sta. 541+53.53 <br />53.53 <br />_949 1;_ m a 11 ,+50 <br />110 40'= " f/ 543+ L. C.= 2 .33.33 <br />86.86 E. C. = Sta. 543 +86.86 . <br />.100_53.53=46.47XX(def. ford ft. of 10° Cur.)=139.41'= <br />2° 191'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Dtf, for 36.86 ft..=1° 501' for a 10° Curve. <br />I.P.An9.23020' " <br />M <br />